Divisibility by 37 . Let the sum of two three-digit numbers be divisible by 37. Prove that the six-digit number obtained by concatenating the digits of those numbers is also divisible by 37.
$\overline {abc}$ + $\overline {def}$ is divisible by 37. Prove $$\overline{abcdef}$$ is divisible by 37.
$$\overline {abc} = 100a + 10b + c$$
$$\overline {def} = 100d + 10e + f$$
then we have
$$\overline {abc}+ \overline {def} = 100a + 10b + c + 100d + 10e + f = 100(a+d) + 10(b+e) + c + f $$
And I'm stuck here. Can anyone help me?
 A: Let $x$ denote $\overline{abc}$.
Let $y$ denote $\overline{def}$.
Hence $1000x+y=\overline{abcdef}$.
Then $\color\red{x+y=37n}\implies1000x+y=999x+\color\red{x+y}=999x+\color\red{37n}=37(27x+n)$.
A: $$\overline{abcdef}=1000\cdot \overline {abc}+\overline {def}$$
$$=999\overline {abc}+\overline {abc}+\overline {def}$$
$$=(37\cdot 27 \cdot \overline {abc})+(\overline {abc}+\overline {def})$$
Hope this helps.
A: Notice, since $\overline {abc}+\overline{def}$ is divisible by $37$ hence, $$(100a+10b+c)+(100d+10e+f)=37\lambda $$ or $$100d+10e+f=37\lambda-(100a+10b+c)\tag 1$$
where, $\lambda$ is some integer 
Now, one should have concatenated number as $$\color{blue}{\overline{abcdef}}=100000a+10000b+1000c+100d+10e+f$$
$$=1000(100a+10b+c)+100d+10e+f$$
setting value from (1), 
$$=1000(100a+10b+c)+37\lambda-(100a+10b+c)$$
$$=999(100a+10b+c)+37\lambda$$
$$=37(2700a+270b+27c+\lambda)$$
since, $(2700a+270b+27c+\lambda)$ is an integer, $37(2700a+270b+27c+\lambda)$ is divisible by $37$ i.e. $\color{red}{\overline{abcdef}}$ is divisible by $\color{red}{37}$
A: Since $37 \mid 111$, then $37 \mid 999$. Hence, for really large numbers, first cast out $999s$ untill you get a $3$ or less digit number. Then check that number for divisibility by  $37$.
Example: $N = 285566$.
$285566 \to 285 + 566 = 851 \to |851 - 888| = 37$
Hence $37 \mid 285566$
